$y=\dfrac{\sqrt{2x^2}}{\cos(x)}$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos(x)+4x^2\sin(x)}{2\sqrt{2x^2}\cos^2(x)}$ (Choice B) B $\dfrac{\sqrt{\cos(x)}(2x\cos(x)+x^2\sin(x))}{\cos^2(x)\sqrt{2x^2}}$ (Choice C) C $\dfrac{2x\cos(x)+2x^2\sin(x)}{\sqrt{2x^2}\cos^2(x)}$ (Choice D) D $-\dfrac{x}{\sin\sqrt{2x^2}}$
$\dfrac{\sqrt{2x^2}}{\cos(x)}$ is a quotient of a composite function and another function. Let... $u(x)=\sqrt{x}$ $v(x)=2x^2$ $w(x)=\cos(x)$... then $y=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $\dfrac{dy}{dx}$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\dfrac{1}{2\sqrt{x}}$ $v'(x)=4x$ $w'(x)=-\sin(x)$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{\dfrac{1}{2\sqrt{2x^2}}}4x \cos(x)-{\sqrt{2x^2}}(-\sin(x))}{\left[ \cos(x) \right]^2} \\\\ &=\dfrac{4x\cos(x)+4x^2\sin(x)}{2\sqrt{2x^2}\cos^2(x)} \\\\ &=\dfrac{2x\cos(x)+2x^2\sin(x)}{\sqrt{2x^2}\cos^2(x)} \end{aligned}$ In conclusion: $\dfrac{dy}{dx}=\dfrac{2x\cos(x)+2x^2\sin(x)}{\sqrt{2x^2}\cos^2(x)}$